A body is projected from the ground at an ang | vedclass

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(D) For projectile motion, the angle of projection is $	heta = 	an^{-1}(\sqrt{7})$, so $	an 	heta = \sqrt{7}$.At half of the maximum height $(y =

Vedclass · Accepted Answer

$\frac{3}{4}$

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(D) For projectile motion, the angle of projection is $	heta = 	an^{-1}(\sqrt{7})$, so $	an 	heta = \sqrt{7}$.At half of the maximum height $(y = H/2)$, the vertical component of velocity is given by $v_y^2 = u_y^2 - 2g(H/2) = u^2 \sin^2 	heta - gH$.The horizontal component of velocity remains constant: $v_x = u_x = u \cos 	heta$.The speed $v$ at this height is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 	heta + u^2 \sin^2 	heta - gH} = \sqrt{u^2 - gH}$.Since $H = \frac{u^2 \sin^2 	heta}{2g}$, we have $gH = \frac{u^2 \sin^2 	heta}{2}$.Substituting this into the expression for $v^2$:$v^2 = u^2 - \frac{u^2 \sin^2 	heta}{2} = u^2 (1 - \frac{\sin^2 	heta}{2})$.Given $	an 	heta = \sqrt{7}$, we have $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta} = \frac{7}{1 + 7} = \frac{7}{8}$.Thus, $v^2 = u^2 (1 - \frac{7/8}{2}) = u^2 (1 - \frac{7}{16}) = u^2 (\frac{9}{16})$.Since $v = nu$, we have $n^2 = \frac{9}{16}$, which gives $n = \frac{3}{4}$.

$A$ body is projected from the ground at an angle of $\tan^{-1}(\sqrt{7})$ with the horizontal. At half of the maximum height, the speed of the body is '$n$' times the speed of projection. The value of '$n$' is

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